We consider the system merge.

  Alphabet:

    cons : [nat * list] --> list 
    map : [nat -> nat * list] --> list 
    merge : [list * list * list] --> list 
    nil : [] --> list 

  Rules:

    merge(nil, nil, x) => x 
    merge(nil, cons(x, y), z) => merge(y, nil, cons(x, z)) 
    merge(cons(x, y), z, u) => merge(z, y, cons(x, u)) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

We observe that the rules contain a first-order subset:

  merge(nil, nil, X) => X 
  merge(nil, cons(X, Y), Z) => merge(Y, nil, cons(X, Z)) 
  merge(cons(X, Y), Z, U) => merge(Z, Y, cons(X, U)) 

Moreover, the system is orthogonal.  Thus, by [Kop12, Thm. 7.55], we may omit all first-order dependency pairs from the dependency pair problem (DP(R), R) if this first-order part is terminating when seen as a many-sorted first-order TRS.

According to the external first-order termination prover, this system is indeed terminating:

 || proof of resources/system.trs
 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
 || 
 || 
 || Termination w.r.t. Q of the given QTRS could be proven:
 || 
 || (0) QTRS
 || (1) QTRSRRRProof [EQUIVALENT]
 || (2) QTRS
 || (3) RisEmptyProof [EQUIVALENT]
 || (4) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    merge(nil, nil, %X) -> %X
 ||    merge(nil, cons(%X, %Y), %Z) -> merge(%Y, nil, cons(%X, %Z))
 ||    merge(cons(%X, %Y), %Z, %U) -> merge(%Z, %Y, cons(%X, %U))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(cons(x_1, x_2)) = 1 + x_1 + x_2
 ||    POL(merge(x_1, x_2, x_3)) = 2 + 2*x_1 + 2*x_2 + x_3
 ||    POL(nil) = 2
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    merge(nil, nil, %X) -> %X
 ||    merge(nil, cons(%X, %Y), %Z) -> merge(%Y, nil, cons(%X, %Z))
 ||    merge(cons(%X, %Y), %Z, %U) -> merge(%Z, %Y, cons(%X, %U))
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || R is empty.
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (3) RisEmptyProof (EQUIVALENT)
 || The TRS R is empty. Hence, termination is trivially proven.
 || ----------------------------------------
 || 
 || (4)
 || YES
 || 
We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs.

After applying [Kop12, Thm. 7.22] to denote collapsing dependency pairs in an extended form, we thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:

    0] map#(F, cons(X, Y)) =#> F(X)   
    1] map#(F, cons(X, Y)) =#> map#(F, Y)   

  Rules R_0:

    merge(nil, nil, X) => X 
    merge(nil, cons(X, Y), Z) => merge(Y, nil, cons(X, Z)) 
    merge(cons(X, Y), Z, U) => merge(Z, Y, cons(X, U)) 
    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 

Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_0, minimal, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70].  Thus, we must orient:

  map#(F, cons(X, Y)) >? F(X) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  merge(nil, nil, X) >= X 
  merge(nil, cons(X, Y), Z) >= merge(Y, nil, cons(X, Z)) 
  merge(cons(X, Y), Z, U) >= merge(Z, Y, cons(X, U)) 
  map(F, nil) >= nil 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  cons = \y0y1.1 + y0 + y1 
  map = \G0y1.2y1 + y1G0(y1) 
  map# = \G0y1.3 + y1 + y1G0(y1) 
  merge = \y0y1y2.y0 + y1 + y2 
  nil = 0 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 4 + x1 + x2 + F0(1 + x1 + x2) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) > F0(x1) = [[_F0(_x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 4 + x1 + x2 + F0(1 + x1 + x2) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) > 3 + x2 + x2F0(x2) = [[map#(_F0, _x2)]] 
  [[merge(nil, nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[merge(nil, cons(_x0, _x1), _x2)]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[merge(_x1, nil, cons(_x0, _x2))]] 
  [[merge(cons(_x0, _x1), _x2, _x3)]] = 1 + x0 + x1 + x2 + x3 >= 1 + x0 + x1 + x2 + x3 = [[merge(_x2, _x1, cons(_x0, _x3))]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + 2x1 + 2x2 + F0(1 + x1 + x2) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) >= 1 + 2x2 + x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_0, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.